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How would I find the series expansion $\displaystyle\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$ so that it will turn into an infinite power series again??

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    $\begingroup$ This infinite power series looks like it is counting some partitions of integers. What I would do to find the answer is to calculate the coefficient of the first few terms then use OEIS to find the combinatorial meaning. $\endgroup$ – clark Dec 21 '14 at 2:27
  • $\begingroup$ @clark: I'm not exactly sure how to find OEIS, but I did post the original question here: text $\endgroup$ – Mathy Person Dec 21 '14 at 2:28
  • $\begingroup$ @vadim123 typo, i meant to say i don't know how to use* OEIS $\endgroup$ – Mathy Person Dec 21 '14 at 2:46
  • $\begingroup$ here is the OEIS results: oeis.org/A000041 $\endgroup$ – Math-fun Dec 22 '14 at 11:16
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If you set

$$f(x)=\prod_{n=0}^\infty (1-x^n)^{-1}=\sum_{n=0}^\infty p(n)x^n$$

we see that yours is just

$$f(x^2)=\sum_{n=0}^\infty p(n)x^{2n}.$$

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$$\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}\cdots$$ $$\frac{1}{1+x^k}=1-x^k+x^{2k}-x^{3k}\cdots$$

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  • $\begingroup$ Would that result in....$(1-x+x^2-x^3...)(1+x+x^2+x^3+....)(1-x^2+x^4-x^6....)(1+x^2+x^4+x^6....)(1-x^3+x^6-x^9....)(1+x^3+x^6+x^9+....)....$? $\endgroup$ – Mathy Person Dec 21 '14 at 2:53
  • $\begingroup$ Is there an alternative to find the truncation of $x^{10}$ from that without doing messy expansions? $\endgroup$ – Mathy Person Dec 21 '14 at 3:09
  • $\begingroup$ Yes, and yes. Use the other product I gave you in the solution to your other question. $\endgroup$ – vadim123 Dec 21 '14 at 3:13
  • $\begingroup$ Which product? In the comment, or in your answer? $\endgroup$ – Mathy Person Dec 21 '14 at 3:19
  • $\begingroup$ Are you referring to $$\prod_{k>0}1-x^{2k-1}=\prod_{k>0}\frac{1}{1+x^k}$$? $\endgroup$ – Mathy Person Dec 21 '14 at 3:30
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I suppose that you can make the expresion shorter using $(1+a)(1-a)=1-a^2$ so $$A=\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}=\frac{1}{(1−x^2)(1−x^4)(1-x^6)\cdots}$$ and now use the fact that $$\frac{1}{1-y}=\sum_{i=0}^{\infty}y^i$$ and replace successively $y$ by $x^2$, $x^4\cdots$,$x^{2n}$ before computing the overall product.

Doing so, for a large number of terms, you should arrive to $$A=1+x^2+2 x^4+3 x^6+5 x^8+7 x^{10}+11 x^{12}+15 x^{14}+22 x^{16}+30 x^{18}+42 x^{20}+56 x^{22}+77 x^{24}+101 x^{26}+135 x^{28}+176 x^{30}+O\left(x^{31}\right)$$ and, as mentioned earlier, the coefficients correspond to sequence $\text{A00041}$ of $\text{EOIS}$ where $a_n$ is the number of partitions of $n$ (the partition numbers).

Probably off-topic, for an infinite number of terms, $$A=\frac{1}{\left(x^2;x^2\right){}_{\infty }}$$ where appears the q-Pochhammer symbol and which, for sure, leads to the same expansion.

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